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anonymous

  • 5 years ago

dy/dx of cos(x+y)=x.

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  1. anonymous
    • 5 years ago
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    I came up with - (1+sin(x+y))/(sin(x+y)) but it doesn't really seem right? If you can help, please do. :)

  2. anonymous
    • 5 years ago
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    \[\cos(x+y)=x\] \[\cos^{-1} x=x+y\] \[y=\cos^{-1} x-x\] \[dy/dx=(\cos^{-1} x-x)\prime\]

  3. anonymous
    • 5 years ago
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    \[dy/dx=-1/\sqrt{1-x^2}-1\]

  4. anonymous
    • 5 years ago
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    i hope i am not doing any mistake...

  5. anonymous
    • 5 years ago
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    if this is right, fan me plz.

  6. anonymous
    • 5 years ago
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    You probably used implicit differentiation heist. If you did, then you'd get your answer (and I got the same thing). You could also check it here. http://www.wolframalpha.com/input/?i=derivative+of+cos%28x+%2B+y+%29+%3D+x eDadou, I'm not sure why your method doesn't give the same answer. I think it might have to do with taking the arccosine of both sides, since that then restricts the domain over which x can vary.

  7. anonymous
    • 5 years ago
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    yeah probably.

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